1
(Due Date: 02/06/2012)
Homework # 1, PHYS 121
Spring 2012
Mastering Physics: P2.17, P2.22, P2.19, P2.25, P2.37, P2.67, P2.72
P2.17.
Prepare:
The graph in Figure P2.17 shows distinct slopes in the time intervals: 0 – 1s, 1 s – 2 s, and
2 s – 4 s.
We can thus obtain the velocity values from this graph using
v
=
x
/
t
.
Solve:
(a)
(b)
There is only one turning point. At
t
= 2 s the velocity changes from +20 m/s to
10 m/s, thus reversing the direction of
motion. At
t
= 1 s, there is an abrupt change in motion from rest to +20 m/s, but there is no reversal in motion.
Assess:
As shown above in
(a)
, a positive slope must give a positive velocity and a negative slope must yield a negative
velocity.
P2.19.
Prepare:
Please refer to Figure P2.19. Since displacement is equal to the area under the velocity graph between
t
i
and
t
f
, we can find the car’s final position from its initial position and the area.
Solve:
(a)
Using the equation
x
f
=
x
i
+ area of the velocity graph between
t
i
and
t
f
,
2 s
3 s
4 s
3 s
10 m
area of trapezoid between 0 s and 2 s
1
10 m
(12 m/s
4 m/s)(2 s)
26 m
2
10 m
area of triangle between 0 s and 3 s
1
10 m
(12 m/s)(3 s)
28 m
2
area between 3 s and 4 s
1
28 m
( 4 m/s
2
x
x
xx
)(1 s)
26 m
(b)
The car reverses direction at
t
= 3 s, because its velocity becomes negative.
Assess:
The car starts at
x
i
= 10 m at
t
i
= 0. Its velocity decreases as time increases, is zero at
t
= 3 s, and then becomes
negative. The slope of the velocityversustime graph is negative which means the car’s acceleration is negative and a
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 Acceleration, Work, Velocity, m/s, TI

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